How was the presumed age of the Universe determined?

Apr 23, 2018

Using Hubble's law.

Explanation:

Hubble's law states that the further away a galaxy is, the faster it is moving:

$v \propto d$

Because of this law, if it is extrapolated backward, it is implied that everything in the universe was once concentrated at one point- supporting the idea of the big bang and also makes it possible to estimate how long ago it was when everything was in one place- i.e the birth of the universe.

However, this is not using SI units, but rather the units for velocity is $k m {s}^{-} 1$ and distance is measured in Mega-parsecs $M P c$.

This equation, being linear must have a constant- Hubble's constant:

${H}_{o}$

Making the equation:

$v = {H}_{o} d$

The value of the constant varies throughout the universe but on a VERY rough estimate we can say that the value of the constant is :

${H}_{o} = 70 k m {s}^{-} 1 M P {c}^{-} 1$

This constant allows us to estimate the age of the universe using the equation:

$\frac{1}{H} _ o \approx T$

However, the Hubble constant must be transferred into SI units for this to work...

So let's find the value of the constant:
$70 k m {s}^{-} 1 = 70 000 m {s}^{-} 1$

1 parsec= 3.26 ly= $3.08 \times {10}^{16} m$

1 Mega-parsec(1 million parsecs)=$\left(3.08 \times {10}^{16} \times {10}^{6} m\right)$

${H}_{o} = 70 k m {s}^{-} 1 M P {c}^{-} 1 = \frac{70 \times {10}^{3}}{{10}^{6} \times 3.08 \times {10}^{16}} = 2.27 \times {10}^{-} 18$

Inverse it to find Hubble time:

$\frac{1}{2.27 \times {10}^{-} 18} \approx T$

$T \approx 4.4 \times {10}^{17} s$

$T \approx 1.4 \times {10}^{10}$ years

Which is almost 14 billion years, but scientists often say that the value is 13.8 with an uncertainty of $\pm 0.2$ billion years, so we are within the acceptable range.

And that is basically how the age of the universe can be estimated.